Answer: Choice B
Discriminant = -4; no real solutions
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Explanation:
The given quadratic is x^2-4x+5, which is the same as 1x^2-4x+5
We see that
a = 1
b = -4
c = 5
The discriminant is
d = b^2 - 4ac
d = (-4)^2 - 4(1)(5)
d = 16 - 20
d = -4
The discriminant is -4. The negative discriminant means that we have no real solutions. Instead, the two solutions are nonreal complex values in the form [tex]a+bi[/tex] where [tex]i = \sqrt{-1}[/tex]
Answer:
The discriminant is -4, so the equation has no real solutions.
Step-by-step explanation:
The discriminant of the quadratic polynomial [tex]ax^2 + bx + c[/tex] with [tex]a\not= 0[/tex] is:
[tex]\Delta = b^2 -4ac[/tex]
So our variables are:
[tex]a = 1[/tex]
[tex]b = -4[/tex]
[tex]c = 5[/tex]
Replacing in the formula for find the discriminant we have:
[tex]4^2 - 4(1)(5) = 16 -20 = -4[/tex]
Now if you remember the quadratic formula is the next
[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
The discriminant appear inside a square root that's meaning that is impossible have negative values inside an even root (in the real numbers).
If we try to find the square root of negative 4 that's meaning find a number that multiply by itself give me -4.
[tex]2 \cdot 2 = 4\\-2 \cdot -2 = 4[/tex]
How you can see is impossible get the solution of a negative base (the base is the number inside the root) when is an even root. In a general case this would be of the next form:
[tex]\sqrt[2n]{-a} = \text{not found}[/tex]
For this we can assure that doesn't exist solutions in the real numbers, so the equation has no real solutions.
